Probability & Certainty: Intro Probability & Certainty

Probability & Certainty: Intro Overview History of probability theory Some basic probability theory –Calculating simple probabilities –Combining mutually-exclusive probabilities –Combining independent probabilities –More complex probabilities –Calculating conditional probabilities Bayes’ Theorem, and why we should care about it A test case: The ‘Let’s Make A Deal!’ problem

Probability & Certainty: Intro History of probability theory Compress all of human history (350K years) in one 24- hour day: –the first recorded general problem representation (geometry, invented by Thales of Miletus about 450 B.C.) would have appeared only 9 minutes and 30 seconds ago – the first systematic large-scale collection of empirical facts (Tycho Brahe’s collection of astronomical observations) would have appeared only a minute and a half ago –the first mathematical equation which was able to predict an empirical phenomena (Newton’s 1697 equation for planetary motion) would have appeared only one minute and twelve seconds ago –Probability theory appeared between 1654 (a minute and a half ago) and 1843 (34 seconds ago).

Probability & Certainty: Intro History of probability theory The emergence of elementary probability theory in the 1650s met with enormous resistance and lack of comprehension when it was first introduced, despite its formal character, its utility, and (what we now recognize as) its simplicity. The difficult points (Margolis, 1993) were philosophical rather than mathematical: the notions that one could count possibilities that had never actually existed, and that order could be obtained from randomness.

Probability & Certainty: Intro Basic probability theory: Example 1 A boring standard example: How likely is it that we will throw a 6 with one dice? Basic principle: The probability of any particular event is equal to the ratio of the number of ways the event can happen over the number of ways anything can happen (= the number of ways the event can fail to happen + the number of ways it can happen).

Probability & Certainty: Intro Basic probability theory: Example 2 How likely is it that we will throw a 7 with two die? There is more than one way for the event to happen: 6 + 1, 1+6, 5 + 2, 2+5, 3 + 4, = 6 ways There are 36 (6 x 6) ways for anything to happen So the probability of 7 with two die is 6/36 or 1/6.

Probability & Certainty: Intro Basic probability theory: Example 3 We roll the die 4 times, and never get a seven. What is the probability that we’ll get on the 5th roll? Independent events are events that don’t effect each other’s probability. Since the every roll is independent of every other, the odds are still 1/6.

Probability & Certainty: Intro Basic probability theory: Example 4 We roll the die twice and get a seven both times. What is the probability of that? To combine probabilities of independent events, multiply the odds of each event. The odds of each 7 are 1/6, so the odds of two rolls of 7 in a row are 1/6 x 1/6 = 1/36.

Probability & Certainty: Intro Basic probability theory: Example 5 We roll the die. What are the odds are getting either a 7 or a 2? Now the events are mutually exclusive: if one happens, the other cannot. To combine probabilities of mutually exclusive events, add them together. 6/36 [odds of 7] + 1/36 [odds of 2] = 7/36

Probability & Certainty: Intro Basic probability theory: Example 6.1 We roll the die twice. What are the odds that we get at least one 7 from the two rolls? Can we just add the probabilities of getting a 7 on each roll? No: because the events are not mutually exclusive anymore if we could, we’d be above 100%- guaranteed a 7- after 6 rolls! Where could the guarantee come from?

Probability & Certainty: Intro Basic probability theory: Example 6.2 We roll the die twice. What are the odds that we get at least one 7 from the two rolls? Can we just multiply the probabilities of getting a 7 on each roll? No: because the events are not independent (Why not?) And 1/6 x 1/6 is less than 1/6- so we’d have smaller chance with two rolls than with one!

Probability & Certainty: Intro Basic probability theory: Example 6.3 We roll the die twice. What are the odds that we get at least one 7 from the two rolls? We can turn part of the problem into a problem of mutual exclusivity by asking: what are the odds of there being exactly one seven out of two rolls? one way is to roll 7 first, but not second - the odds of this are 1/6 * 5/6 (independent events) = the odds of rolling 7 second are 5/6 * 1/6 (independent events) = since these two outcomes are mutually exclusive, we can add them to get = 0.277

Probability & Certainty: Intro Basic probability theory: Example 6.4 We roll the die twice. What are the odds that we get at least one 7 from the two rolls? Now we need to know: what are the odds of there being two sevens out of two rolls? We already know: it’s 1/36 = 0.03 So: the odds that we get at least one 7 is the odds of two 7s + the odds of one 7 (mutually exclusive events) = = 0.3

Probability & Certainty: Intro Basic probability theory: The generalization Does anyone know how to generalize this calculation, so we can easily calculate the odds of an event of probability p happening r times out of n tries, for any values of p,r, and n? The binomial theorem: Tune in for the exciting next class!

Probability & Certainty: Intro Basic probability theory: Conditional probability What if a dice is biased so that it rolls 6 twice as often as every other number? How can we deal with ‘uneven’ base rates? Why should we care? –Because real life uses biased dice –eg. the conditional probability of being schizophrenic, given that a person has an appointment with a doctor who specializes in schizophrenia, is quite different from the unconditional probability that a person has schizophrenia (the base rate)

Probability & Certainty: Intro Basic probability theory: Conditional probability A particular disorder has a base rate occurrence of 1/1000 people. A test to detect this disease has a false positive rate of 5%. Assume that the test diagnoses correctly every person who has the disease. What is the chance that a randomly selected person with a positive result actually has the disease? [Take a guess.] Harvard Med School estimates: About half said 95%. Average response was 56%. Only 16% gave the correct answer.

Probability & Certainty: Intro Basic probability theory: Conditional probability A particular disorder has a base rate occurrence of 1/1000 people. A test to detect this disease has a false positive rate of 5%. Assume that the test diagnoses correctly every person who has the disease. What is the chance that a randomly selected person with a positive result actually has the disease? Conditional probability P(A|B) = P(A,B) / P(B) = P(A and B)/P(B) Let A = Has the disorder and B = Has a positive test result In 10,000 people, P(A and B) = P(B) = Chance that a randomly selected person with a positive result actually has the disease= * 1000 = * 1000 (false positives) + 10 (true positives) = /510 = 0.02 or just a 2% chance

Probability & Certainty: Intro A generalization: Bayes’ theorem P(A|B) = P(B|A) P(A) / P(B) Proof: By definition, (1.) P(A|B) = P(A,B) / P(B) (2.) P(B|A) = P(A,B) / P(A) (3.) P(B|A) P(A) = P(A,B) [Multiply (2.) by P(A)] (4.) P(A|B) P(B) = P(B|A) P(A) [Substitute (1.) in (3.)] (5.) P(A|B) = P(B|A) P(A) / P(B)

Probability & Certainty: Intro Let’s try Bayes’ theorem P(A|B) = P(B|A) P(A) / P(B) Let: P(A) = Probability of disease = P(B) = Probability of positive test = = P(B|A) = Probability of positive test given disease = 1 Then: P(A|B) = P(B|A) P(A) / P(B) = (1 x 0.001) / (0.0501) = 0.02

Probability & Certainty: Intro The Notorious 3-Curtain (‘Let’s Make A Deal’) Problem! Three curtains hide prizes. One is good. Two are not. You choose a curtain. The MC opens another curtain. It’s not good. He gives you the chance to stay with our first choice, or switch to the remaining unopened curtain. Should you stay or switch, or does it matter?

Probability & Certainty: Intro P(A|B) = P(B|A) P(A) / P(B) Assume you choose A and switch to the unopened one Let:P(A) = Probability it is behind the unopened one (B or C): 2/3 P(B) = Probability it is not behind A: 2/3 P(B|A) = Probability it is not behind A, given that it is behind the unopened one (B or C): 1 Then: P(A|B) = Probability it is behind the unopened one (B or C), given that it is not behind A = P(B|A) P(A) / P(B) = (1 x 2/3) / 2/3 = 1= 100% of the time. But it is given that it is not behind A 2/3 of the time, so 2/3 of the time we can be certain of winning if we switch!